Question

When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the diameter of this resistor is now 1/3 the original diameter, the current will be?

Answer 1

Answer:

Current would decrease 9 times.

Explanation:

Assume all others (potential difference U, resistivity ρ, and length L) are the same, only change in the diameter, we would have the following ratio

$\frac{I_1}{I_2} = \frac{U/R_1}{U/R_2} = \frac{U}{U}\frac{R_2}{R_1} = \frac{\rho L/A_2}{\rho L / A_1} = \frac{\rho L}{\rho L}\frac{A_1}{A_2} = \frac{\pi d_1^2/4}{\pi d_2^2/4} = \frac{4\pi}{4\pi}\left(\frac{d_1}{d_2}\right)^2 = 3^2 = 9$

$I_2 = I_1/9$

So the current would decrease 9 times.